Question: You have found the following ages (in years) of all 6 bears at your local zoo: $ 18,\enspace 2,\enspace 2,\enspace 23,\enspace 21,\enspace 2$ What is the average age of the bears at your zoo? What is the variance? You may round your answers to the nearest tenth.
Solution: Because we have data for all 6 bears at the zoo, we are able to calculate the population mean $({\mu})$ and population variance $({\sigma^2})$ To find the population mean , add up the values of all $6$ ages and divide by $6$ $ {\mu} = \dfrac{\sum\limits_{i=1}^{{N}} x_i}{{N}} = \dfrac{\sum\limits_{i=1}^{{6}} x_i}{{6}} $ $ {\mu} = \dfrac{18 + 2 + 2 + 23 + 21 + 2}{{6}} = {11.3\text{ years old}} $ Find the squared deviations from the mean for each bear. Age $x_i$ Distance from the mean $(x_i - {\mu})$ $(x_i - {\mu})^2$ $18$ years $6.7$ years $44.89$ years $^2$ $2$ years $-9.3$ years $86.49$ years $^2$ $2$ years $-9.3$ years $86.49$ years $^2$ $23$ years $11.7$ years $136.89$ years $^2$ $21$ years $9.7$ years $94.09$ years $^2$ $2$ years $-9.3$ years $86.49$ years $^2$ Because we used the population mean $({\mu})$ to compute the squared deviations from the mean , we can find the variance $({\sigma^2})$ , without introducing any bias, by simply averaging the squared deviations from the mean $ {\sigma^2} = \dfrac{\sum\limits_{i=1}^{{N}} (x_i - {\mu})^2}{{N}} $ $ {\sigma^2} = \dfrac{{44.89} + {86.49} + {86.49} + {136.89} + {94.09} + {86.49}} {{6}} $ $ {\sigma^2} = \dfrac{{535.34}}{{6}} = {89.22\text{ years}^2} $ The average bear at the zoo is 11.3 years old. The population variance is 89.22 years $^2$.